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解:
(1)证明:由b/c = -(√3cosB-1)/(√3cosC-1)得:
sinB/sinC = -(√3cosB-1)/(√3cosC-1)
sinB(√3cosC-1)+sinC (√3cosB-1)=0
sinB.√3cosC+sinC .√3cosB = sinB+sinC
√3sin(B+C)=sinB+sinC
√3sinA = sinB+sinC
√3a=b+c
(2)由余弦定理:cosA = (b²+c²-a²)/(2bc)
=[b²+c²-(1/3)(b+c)²]/(2bc)
=[(2/3)b²+(2/3)c²-(2/3)bc]/(2bc)
=(1/3)(b/c)+(1/3)(c/b)-1/3
≥ (1/3).2√[(b/c).(c/b)]-1/3
=1/3
当且仅当b=c时,等号成立
得cosA最小值1/3
(1)证明:由b/c = -(√3cosB-1)/(√3cosC-1)得:
sinB/sinC = -(√3cosB-1)/(√3cosC-1)
sinB(√3cosC-1)+sinC (√3cosB-1)=0
sinB.√3cosC+sinC .√3cosB = sinB+sinC
√3sin(B+C)=sinB+sinC
√3sinA = sinB+sinC
√3a=b+c
(2)由余弦定理:cosA = (b²+c²-a²)/(2bc)
=[b²+c²-(1/3)(b+c)²]/(2bc)
=[(2/3)b²+(2/3)c²-(2/3)bc]/(2bc)
=(1/3)(b/c)+(1/3)(c/b)-1/3
≥ (1/3).2√[(b/c).(c/b)]-1/3
=1/3
当且仅当b=c时,等号成立
得cosA最小值1/3
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